Finding Matrix of a linear operator respecting a basis

Finding Matrix of a linear operator respecting a basis

Postby Guest » Sat Apr 15, 2017 2:50 pm

If fi(the greek letter... a linear operator) belongs to Hom(R3) and has in respect to E(the standard basis of R3) has a matrix matrix [tex]f=
\begin{matrix}
1 & 3 & 0 \\
0 & 1 & 0 \\
3 & 5 & 3 \\
\end{matrix}[/tex] , Find the matrix J of fi respecting the basis A.
[tex]A=\{{a_1=(9,0,9),a_2=(0,3,5),a_3=(3,-3,-1)}\}[/tex]
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Re: Finding Matrix of a linear operator respecting a basis

Postby HallsofIvy » Tue Dec 17, 2019 6:56 pm

In basis A, the vector (9, 0, 9) (in the standard basis) is represented by (1, 0, 0), (0, 3, 5) (in the standard basis) by (0, 1, 0), and (3, -3, -1) by (0, 0, 1). Further, with the matrix [tex]\begin{bmatrix}a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}[/tex] applied to (1, 0, 0) gives (a, d, g), applied to (0, 1, 0) gives (b, e, h), and applied to (0, 0, 1) gives (c, f, i).

That is, we can get the columns of the matrix representation of a linear transformation in a given basis by applying the linear transformation to each of the basis vectors in turn, then writing the result in terms of the basis.

Here, we are told that the linear transformation is given, in the standard basis, by the matrix [tex]\begin{bmatrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 3 & 5 & 3\end{bmatrix}[/tex]. Then [tex]\begin{bmatrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 3 & 5 & 3\end{bmatrix}\begin{bmatrix}9 \\ 0 \\ 9\end{bmatrix}= \begin{bmatrix}9 \\ 0 \\ 54\end{bmatrix}[/tex]. To write that in terms of the basis, we need to find a, b, and c such that a(9, 0, 9)+ b(0, 3, 5)+ c(3, -3, -1)= (9a+ 3c, 3b- 3c, 9a+ 5b- c)= (9, 0, 54) so we have the three equations, 9a+ 3c= 9, 3b- 3= 0, and 9a+ 5b- c= 54 to solve for a, b, and c, forming the first row of the matrix.

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